[[[[[

 A LISP expression that evaluates to itself!

 Let f(x): x -> (('x)('x))

 Then (('f)('f)) is a fixed point.

]]]]]

[Here is the fixed point done by hand:]

(
'lambda(x) cons cons "' cons x nil
           cons cons "' cons x nil
                nil

'lambda(x) cons cons "' cons x nil
           cons cons "' cons x nil
                nil
)

[Now let's construct the fixed point.]

define (f x) let y [be] cons "' cons x nil [y is ('x)        ]
             [return] cons y cons y nil    [return (('x)('x))]

[Here we try f:]

(f x)

[Here we use f to calculate the fixed point:]

(f f)

[Here we find the value of the fixed point:]

eval (f f)

[Here we check that it's a fixed point:]

= (f f) eval (f f)

[Just for emphasis:]

= (f f) eval eval eval eval eval eval (f f)
